# Thermo question - final mass/entropy generation

• goblue88888
And I still don't know thermo.) In summary, the problem involves an initially empty insulated tank that is connected to a line of air at 50 degrees Celsius and 200 kilopascals. The valve is opened until the tank is filled, and then closed. The insulation is removed and a heat sink cools the tank to 30 degrees Celsius. The final mass of air in the tank is 1.51 kilograms. The entropy generated during the cooling process is 0.00484 kilojoules per Kelvin.

## Homework Statement

An insulated and initially empty 0.7m^3 tank is connected by a valve to a line flowing air at 50degreesC and 200kPa. The valve is opened until the tank is filled, and then the valve is closed. The insulation is then removed from the tank, and a 20degreeC heat sink cools the tank until the air inside reaches a temperature of 30degreesC. 1) What is the final mass of air in the tank? 2) How much entropy is generated during the cooling of the tank?

## Homework Equations

1)
PV=mRT
2)
S2-S1=Cp*ln(T2/T1)-R*ln(P2/P1)
m(s2-s1)=Integral(dQ/T)+S-gen

## The Attempt at a Solution

1) I tried figuring out the final temperature, and since V and R are constant, I got P2=P1*T2/T1=200kPa*303K/323K=187.62kPa. I then used this P to get m=PV/RT. So m=(187.62*0.7)/(0.287*323)=1.417kg.. Is this right?
2) Using the first equation, I get 1.004*ln(303/323)-0.287*ln(187.62/200)=-0.0458.. But this is wrong right since it shouldn't be negative?
Then I try putting it into the second equation, 1.417*-0.0458 - Integral(dQ/303)=S-gen.. but I don't know how to do the Integral and dQ part... any help? THANKS!

well since no one else is helping, I'll take a stab, but in the buyer beware category, i know very little about thermo.

I believe its negative because it is cooler, and altho under less pressure, the temperature difference is dominant. It may be that the question is asking about what entropy must have been generated in the surroundings to accomplish this which would then be >=0.0458 depending on efficiency. integral dQ here becomes deltaQ i believe.

In 1), your equation for m contains a final state pressure and an initial state temperature. You can compute the mass in either the initial state or final state - but you can't mix and match numbers between the two. For 2), why not just do it in terms of the heat capacity at constant volume? Why bring pressure in at all? Oh, it would also be nice if you could be more complete about including units on quantities.

Dick said:
In 1), your equation for m contains a final state pressure and an initial state temperature. You can compute the mass in either the initial state or final state - but you can't mix and match numbers between the two. For 2), why not just do it in terms of the heat capacity at constant volume? Why bring pressure in at all? Oh, it would also be nice if you could be more complete about including units on quantities.

Dick and Denver! Thanks so much for responding! I really appreciate it...

For 1).. I accidentally put in T1! Oops!
P2 = P1*T2/T1 = 200kPa*303K/323K = 187.62kPa
So it should be (187.62kPa*0.7m^3)/(0.287kJ/kg-K*303K) = 1.51kg.

2)
Do you mean the equation s2-s1=C*ln(T2/T1)? I have that down as entropy change for a liquid or solid.
For an ideal gas (air), I have
s2-s1 =
Cp (specific heat) * ln(T2/T1) - R * ln(P2/P1)
or
Cv (specific heat) * ln(T2/T1) + R * ln(v2/v1)

So I used the first one since I have P. Is my logic completely off? :\

Well, either your logic is off, or it's been too long since I've done thermodynamics. But look at your second form. v2=v1. ln(1)=0.

Dick said:
Well, either your logic is off, or it's been too long since I've done thermodynamics. But look at your second form. v2=v1. ln(1)=0.

OMG Dick! Congrats on 1000 posts! :) :yuck: :!) Haha..

Anyways.. you're right.. I was using the constant pressure equation (boy, do I feel stupid.. although I already do after putting the wrong T in the first post).. ok.. I totally understand now.. I think.. here's my work -

After I get m = 1.51 kg from part 1 -

I get s2-s1 = Cv*ln(T2/T1) + R*ln(v2/v1) = 0.717kj/kg-K * ln(303K/323K) + R*0
= -0.04583 kJ/kg-K

I also need to get 1Q2. I use 1Q2 = 1W2 + m*(u2-u1) = 0 + 1.51kg*(216.52-230.887) = -21.694 kJ. I got the u2-u1 from interpolating using a chart of ideal-gas properties of air.

Then I plug them into the equation with s-gen. So
s-gen = m(s2-s1) - 1Q2/T-source = 1.51kg * -0.04583 kJ/kg-K - (-21.694 kJ)/(20+273.15) = 0.00484kJ/K!

OMG, I think I did it correctly!

Good job! Then that was a worthwhile 1000th post. Thanks!

## What is the final mass in a thermo question?

The final mass in a thermo question refers to the total mass of a system after a thermodynamic process has occurred. This can be calculated using the law of conservation of mass, which states that mass cannot be created or destroyed, only transferred or converted into different forms.

## What is entropy generation in a thermo question?

Entropy generation is a measure of the increase in disorder or randomness within a system as a result of a thermodynamic process. It is a fundamental concept in thermodynamics and is closely related to the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time.

## How is final mass related to entropy generation?

The final mass in a thermo question can affect the amount of entropy generated in a system. This is because any changes in the mass or composition of a system can impact the energy and heat transfer processes within that system, which in turn can affect the level of disorder or randomness within the system.

## What factors can influence final mass and entropy generation?

Various factors can influence final mass and entropy generation in a system, including changes in temperature, pressure, and composition. The type of thermodynamic process taking place, such as isothermal, adiabatic, or isobaric, can also play a role in determining the final mass and entropy generation.

## How are final mass and entropy generation calculated in a thermo question?

The final mass and entropy generation in a thermo question can be calculated using a combination of thermodynamic equations, such as the first and second laws of thermodynamics, along with knowledge of the system's properties, such as temperature, pressure, and volume. It is important to carefully consider all factors and assumptions when solving for these values in a thermo question.